is there a continuous function that maps (0,1) onto [0,1]?
M mihern New member Joined Jun 5, 2008 Messages 1 Jun 5, 2008 #1 is there a continuous function that maps (0,1) onto [0,1]?
skeeter Elite Member Joined Dec 15, 2005 Messages 3,218 Jun 5, 2008 #2 Re: mapping (0,1) onto [0,1] only one I can think of off the top of my head would have to be defined piece-wise ... is that satisfactory? f(x)=1\displaystyle f(x) = 1f(x)=1 0<x≤14\displaystyle 0 < x \leq \frac{1}{4}0<x≤41 f(x)=−2x+32\displaystyle f(x) = -2x + \frac{3}{2}f(x)=−2x+23 14<x<34\displaystyle \frac{1}{4} < x < \frac{3}{4}41<x<43 f(x)=0\displaystyle f(x) = 0f(x)=0 34≤x<1\displaystyle \frac{3}{4} \leq x < 143≤x<1
Re: mapping (0,1) onto [0,1] only one I can think of off the top of my head would have to be defined piece-wise ... is that satisfactory? f(x)=1\displaystyle f(x) = 1f(x)=1 0<x≤14\displaystyle 0 < x \leq \frac{1}{4}0<x≤41 f(x)=−2x+32\displaystyle f(x) = -2x + \frac{3}{2}f(x)=−2x+23 14<x<34\displaystyle \frac{1}{4} < x < \frac{3}{4}41<x<43 f(x)=0\displaystyle f(x) = 0f(x)=0 34≤x<1\displaystyle \frac{3}{4} \leq x < 143≤x<1
